Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{k^2 + 13k + 40}{k^2 + 6k + 5} \times \dfrac{-3k - 3}{-5k + 35} $
Answer: First factor out any common factors. $y = \dfrac{k^2 + 13k + 40}{k^2 + 6k + 5} \times \dfrac{-3(k + 1)}{-5(k - 7)} $ Then factor the quadratic expressions. $y = \dfrac {(k + 5)(k + 8)} {(k + 5)(k + 1)} \times \dfrac {-3(k + 1)} {-5(k - 7)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac { (k + 5)(k + 8) \times -3(k + 1)} { (k + 5)(k + 1) \times -5(k - 7)} $ $y = \dfrac {-3(k + 5)(k + 8)(k + 1)} {-5(k + 5)(k + 1)(k - 7)} $ Notice that $(k + 5)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {-3\cancel{(k + 5)}(k + 8)(k + 1)} {-5\cancel{(k + 5)}(k + 1)(k - 7)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $y = \dfrac {-3\cancel{(k + 5)}(k + 8)\cancel{(k + 1)}} {-5\cancel{(k + 5)}\cancel{(k + 1)}(k - 7)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $y = \dfrac {-3(k + 8)} {-5(k - 7)} $ $ y = \dfrac{3(k + 8)}{5(k - 7)}; k \neq -5; k \neq -1 $